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Math Help - Area of segment.

  1. #1
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    Area of segment.

    A circle has the center of A and the radius of 5. |BC| = 8; |AC|=|AB|=5
    Find the area of the gray bit of the circle.

    Last edited by mr fantastic; May 8th 2010 at 03:59 PM. Reason: Re-titled.
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    Quote Originally Posted by Kane535 View Post
    A circle has the center of A and the radius of 5. |BC| = 8; |AC|=|AB|=5
    Find the area of the gray bit of the circle.

    Hi Kane535, welcome to MHF.
    Altitude from A on BC is 3.
    So area of ABC = 1/2*8*3 = 12 sq. units
    Angle ABC is given by
    θ = 2*arctan(4/3)
    Area of the sector ABC = 1/2*r^2*θ, where θ is in radians.
    Now find the area of the shaded portion.
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    Quote Originally Posted by Kane535 View Post
    A circle has the center of A and the radius of 5. |BC| = 8; |AC|=|AB|=5
    Find the area of the gray bit of the circle.

    Use the cos rule to find angle A (I will use radians because it's easier)

    \cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{5^2+5^2-8^2}{2 \cdot 5 \cdot 5} = -\frac{7}{25}

    A = \arccos \left(-\frac{7}{25}\right) \approx 1.855


    The area of a sector is given by A = \frac{1}{2}r^2 \theta. In this case \theta is angle A and r=5

    A_s = \frac{1}{2} \cdot 5^2 \cdot \arccos \left(-\frac{7}{25}\right)

    Since this includes the white triangle we must remove it's area which is given by A_t = \frac{1}{2} bc \sin A = \frac{1}{2} \cdot 5^2 = \frac{25}{2}

    A_g = A_s - A_t = \frac{1}{2} \cdot 5^2 \cdot \arccos \left(-\frac{7}{25}\right) - \frac{25}{2}

    Evaluate for the answer, I get an answer of 10.7 \text{  3sf }
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